Angle to Shoot Bow & Arrow at Object 1m Deep in Lake

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Homework Help Overview

The problem involves a scenario where a man is trying to shoot an arrow at a treasure chest located 1 meter deep in a lake. The man’s eyes are positioned 2 meters above ground level, and he is looking at the chest at an angle of 30 degrees below the horizontal. The discussion centers on determining the appropriate angle to shoot the arrow, taking into account the effects of refraction as light passes from air into water.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the application of the law of refraction and the relationship between incident and refracted angles. There is an attempt to clarify the angles involved, particularly how the angle of 30 degrees relates to the vertical normal. Questions arise regarding the use of given heights and the correct substitutions in the refraction formula.

Discussion Status

The discussion is ongoing, with participants providing guidance on the refraction formula and suggesting the need for clearer diagrams. Some participants express uncertainty about their calculations and seek validation of their approaches, indicating a collaborative effort to understand the problem better.

Contextual Notes

Participants note potential confusion regarding the application of the heights provided in the problem and the implications of the angles with respect to the vertical normal. There is also mention of an attachment that requires approval for visibility, which may hinder the discussion of visual aids.

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Homework Statement


A man whose eyes are 2 m above the ground is looking into a clear lake. He sees a treasure chest on the bottom of the lake, which is 1 m deep. His eyes make an angle of 30 degrees below the horizontal as he looks at the chest.

If he wants to shoot the chest with a bow and arrow (so he can pull it to the surface), at what angle should he shoot the arrow?

State the angle with respect to the horizontal.

Homework Equations


I learned the law of refraction was nisin(θi) = nrsin(θr). The angle is with respect to a VERTICAL normal. so \| the angle between those two lines is the angle of the ray going in, |\ that's the angle going out.

i means incidient, r means refracted.


The Attempt at a Solution


I can't get an answer for this, but I'm not sure what I'm doing wrong.

I'm assuming that when the guy is looking at the object, he's seeing the refracted object. The real object is actually in a different spot. So the angle that his eyes make with respect to the vertical should be θr.

If the guy's eyes make an angle of 30 degrees from the horizontal, then the angle with respect to a vertical normal must be 60 degrees.

Then using the equation given I should be able to solve for θi. But I can't, because it gives sin-1 > 1, which is impossible.

Also, I didn't use the heights given to me at all and I think I'm supposed to. What am I doing wrong?
 
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You must have mixed up the substitution into the refraction formula. On your diagram, label the air side 1 and write n1 = 1. On the water side 2, n2 = 1.33. Show your formula with numbers if you need more help!
 
Check out this amaturistic diagram, hope it helps?:confused:
 

Attachments

  • defraction ratio.jpg
    defraction ratio.jpg
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Please tell me is this anywhere close?
 
No one can see your attachment until it is approved.
It would be quicker if you uploaded it to a photo site like photobucket.com and then posted a link here. If you put IMG and /IMG (both in square brackets) around the link, the image will appear right in the post.
 

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